Problem Description

Given a string s and an integer k, return the total number of substrings of s where at least one character appears at least k times.

Intuition

To solve this problem, we use a sliding window technique to efficiently count the substrings meeting the condition. The idea is to maintain a window (a segment of the string) where we can count the characters using a frequency counter. We adjust the window’s left and right edges as we iterate through the string.

1. Sliding Window Approach: We iterate through the string using a pointer (r) for the right end. For each position, we add the character's count to a counter.

2. Condition Check: If the character count in the window reaches k or more, it means that any substring starting from the beginning of the string to the left end (l-1) and ending at the current position (r) will meet the condition.

3. Adjust Window: We then increment the left end (l) to decrease the count of the character at the left end until no character in the window has a count of k or more, ensuring that we only count substrings satisfying the problem requirements.

4. Accumulate Result: For each valid window, add the number of valid starting points, l, to the answer.

By maintaining this window dynamically, we ensure that each examined substring is counted, resulting in an efficient calculation.

Learn more about Sliding Window patterns.

Solution Approach

The solution uses a Sliding Window approach to efficiently count the desired substrings in the given string s. Here is the step-by-step process:

1. Initialize Variables:

   • cnt: A Counter from the collections module to track the frequency of each character within the current window.
   • ans: To store the total count of valid substrings. Initialized to 0.
   • l: The left end of the sliding window. Initialized to 0.

2. Iterate Over String:

   • Use a loop to iterate over each character c in the string s. For each character, perform the following:
     • Increase the count of c in cnt by 1 to reflect its presence in the current window.

3. Adjust Left End of Window:

   • While the count of the current character c in cnt is greater than or equal to k, it means any starting point for a substring from 0 to l-1 with the ending point at the current position satisfies the condition.
     • To remove such overlapping substrings, decrease the count of the character s[l] in cnt by 1 (effectively sliding the window to the right by moving its left edge l).
     • Increment l to reflect the new position of the left edge.

4. Count Valid Substrings:

   • For each valid window (where no character frequency is k or more), add l to ans, representing all valid substrings with the current character as the right endpoint.

5. Return Result:

   • After completing the iteration over the string, ans will hold the total count of substrings satisfying the given condition. Return ans as the final result.

This method efficiently manages the character counts and keeps adjusting the window's boundaries as required, allowing for optimal calculation of valid substrings within the given constraints.

Example Walkthrough

Let's use the string s = "aabab" and k = 2 to illustrate the sliding window approach for counting valid substrings where at least one character appears at least k times.

1. Initialize Variables:

   • Start with an empty dictionary cnt to keep track of character frequencies.
   • Set ans = 0 to store the count of valid substrings.
   • Initialize l = 0 to mark the left edge of the sliding window.

2. Iterate Over String:

   • Begin iterating with r = 0 (where r is the right edge of the window):

     • Add s[0] = 'a' to the cnt dictionary: {'a': 1}.
     • Since no character appears k times, continue to the next character.

   • Move to r = 1:

     • Add s[1] = 'a' to cnt: {'a': 2}.
     • Now, 'a' appears at least k times (2 times).
     • We can count substrings ending at r = 1 starting from any point 0 to l, so add l + 1 = 1 to ans.

3. Adjust Left End of Window:

   • Move r = 2:

     • Add s[2] = 'b' to cnt: {'a': 2, 'b': 1}.
     • 'a' still appears k times, so count the same substrings: ans += l + 1 = 2.

   • Progress to r = 3:

     • Add s[3] = 'a' to cnt: {'a': 3, 'b': 1}.
     • Again, 'a' appears 3 times, which is more than k. Hence, count substrings: ans += l + 1 = 3.

4. Continue Adjusting:

   • Move to r = 4:
     • Add s[4] = 'b' to cnt: {'a': 3, 'b': 2}.
     • Both 'a' and 'b' appear at least k times. Now we adjust the window by incrementing l and changing cnt:
       • First, decrease cnt['a'] by removing s[l] = 'a': {'a': 2, 'b': 2}, then increment l = 1.
       • After adjustment, no extra valid occurences to count.

Finally, ans accumulates to 8, which is the total number of substrings where at least one character appears at least k times. Return ans.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string s. This is because each character in the string is processed exactly once by both the loop that iterates through s and the inner while loop, due to the window sliding mechanism.

The space complexity is O(|Σ|), where |Σ| is the size of the character set. In this case, the character set is the lowercase English letters, so |Σ| = 26. This space is used by the Counter to store the frequencies of characters currently in the sliding window.

Learn more about how to find time and space complexity quickly.